digital communication
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Introduction to Digital Communications
Midterm
3:30AM ~ 6:30PM, 4/26/2016
1. (2+3+2+3=10 points)
Given the carrier frequency fc , symbol energy E, and symbol duration T, write
down the mathematical formats of the transmitted signals of
A. M-ary PSK.
B. M-ary QAM square constellations (given the in-phase symbol ai and
quadrature symbol bi, and the energy E0 of the signal with the lowest
amplitude.)
C. M-ary FSK (supposing that fc = nc /2T).
D. Carrierless amplitude and phase modulation (CAP) (given a pulse shaping
function of g(t), with g(t) containing no symbol energy)
[A]:
A. ๐ ๐(๐ก) = โ2๐ธ
๐cos (2๐๐๐๐ก +
2๐
๐(๐ โ 1)), 0 โค ๐ก < ๐, ๐ = 1,โฆ ,๐
B. ๐ ๐(๐ก) = โ2๐ธ0
๐๐๐ cos 2๐๐๐๐ก โโ
2๐ธ0
๐๐๐ sin 2๐๐๐๐ก , 0 โค ๐ก โค ๐ with ai and bi are
integers. Define ๐ฟ = โ๐, then we have
{๐๐, ๐๐} = [
(โ๐ฟ + 1, ๐ฟ โ 1) (โ๐ฟ + 3, ๐ฟ โ 1)(โ๐ฟ + 1, ๐ฟ โ 3) (โ๐ฟ + 3, ๐ฟ โ 3)
โฆ (๐ฟ โ 1, ๐ฟ โ 1)
โฆ (๐ฟ โ 1, ๐ฟ โ 3)โฎ โฎ
(โ๐ฟ + 1,โ๐ฟ + 1) (โ๐ฟ + 3,โ๐ฟ + 1)โฑ โฎโฆ (๐ฟ โ 1,โ๐ฟ + 1)
]
C. ๐ ๐(๐ก) = {โ2๐ธ
๐cos (
๐
๐(๐๐ + ๐)๐ก) , 0 โค ๐ก โค ๐
0, elsewhere
, i = 1,2,โฆ,M,
D. ๐ (๐ก) = Re{โ (๐๐ + ๐๐๐)๐(๐ก โ ๐๐)exp(๐2๐๐๐๐ก) โ๐=โโ }
Remember to write down your id number and your name.
Please provide detailed explanations/derivations in your answers. Correct answer without any explanations
will not be given any credits. However, wrong answers with correct reasoning will have partial credits.
2. (2+3+5=10 points)
Following the previous problem, given the carrier frequency fc , symbol energy E,
and symbol duration T, provide the basis functions and the message points of the
following modulations
A. M-ary PSK.
B. BFSK, also called Sundeโs FSK, (supposing that fc = (f1 + f2)/2).
C. Carrierless amplitude and phase modulation (CAP) (given a pulse shaping
function of g(t)).
[A]:
A. The orthonormal bases for M-ary PSK ae
๐1(๐ก) = โ2
๐cos 2๐๐๐๐ก 0 โค ๐ก โค ๐
๐2(๐ก) = โโ2
๐sin 2๐๐๐๐ก 0 โค ๐ก โค ๐
The message points are given by.
๐ฅ๐ผ = โ๐ธcos (2๐
๐(๐ โ 1)) ๐ฅ๐ = โ๐ธsin (
2๐
๐(๐ โ 1))
B. The orthonormal bases functions for Sundeโs FSK are
๐๐(๐ก) = {โ
2
๐๐cos(2๐๐๐๐ก), 0 โค ๐ก < ๐๐
0, elsewhere
The two message points on are
๐บ1 = [โ๐ธ๐0
] and ๐๐ = [0
โ๐ธ๐],
C. The passband in-phase pulse is ๐(๐ก) = ๐(๐ก) cos(2๐๐๐๐ก), and
the passband quadrature pulse is ๏ฟฝฬ๏ฟฝ(๐ก) = ๐(๐ก) sin(2๐๐๐๐ก).
The message points are {๐๐, ๐๐}
3. (4+8+4+4=20 points)
Given the carrier frequency fc , symbol energy Eb, and symbol duration Tb, for
Minimum shift keying (MSK).
A. Provide the mathematical format of the transmitted signal
B. Provide the basis functions and the message points
C. Show the phase trellis of the MSK signals for a sequence of binary waves:
1 1 0 1 0 0 1 0
D. For Gaussian-filtered MSK (GMSK), given the pulse shaping function
๐(๐ก) =1
2[erfc (๐โ
2
log2๐๐๐ (
๐ก
๐๐โ
1
2)) โ erfc (๐โ
2
log2๐๐๐ (
๐ก
๐๐+
1
2))]
with W being the 3 dB baseband bandwidth of the pulse shaping filter.
Explain how GMSK might affect the phase trellis for the cases of ๐๐๐=
and ๐๐๐
[A]:
A. ๐ (๐ก) = โ2๐ธ๐
๐๐cos ๐(๐ก) cos(2๐๐๐๐ก) โ โ
2๐ธ๐
๐๐sin ๐(๐ก) sin(2๐๐๐๐ก)
With ๐(๐ก) = ๐(0) ยฑ๐
2๐๐๐ก, 0 โค ๐ก โค ๐๐
B. The in-phase term for โ๐๐ โค ๐ก โค ๐๐ is
๐1(๐ก) = โ2
๐๐๐๐๐ (
๐
2๐๐๐ก) ๐๐๐ 2๐๐๐๐ก
The message point is ๐ 1 = โซ ๐ (๐ก)๐1(๐ก)๐๐โ๐๐
๐๐ก = โ๐ธ๐cos ๐(0).
On the other hand, the quadrature term for 0 โค ๐ก โค 2๐๐ is
๐2(๐ก) = โ2
๐๐๐ ๐๐ (
๐
2๐๐๐ก) ๐ ๐๐ 2๐๐๐๐ก
The message point is ๐ 2 = โซ ๐ (๐ก)๐2(๐ก)2๐๐0
๐๐ก = โโ๐ธ๐sin ๐(๐๐)
D. Phase transitions in the phase diagram become smoother.
4. (5+5+10+5=25 points)
Given that the bit energy is Eb and the noise density is N0,
A. Show the observation vectors of the received QPSK signals, and accordingly
provide the decision rule and its corresponding decision boundaries in the
signal space of QPSK
B. Suppose the input dibits are Gray-encoded, derive the bit error rate of
QPSK given that 1
โ๐โซ exp[โ๐ง2]๐๐งโ
๐ฅ=
1
2erfc(๐ฅ)
C. Show the observation vectors of the received MSK signals, and accordingly
provide the detection rule and its corresponding decision boundaries in the
signal space of MSK.
D. Derive the approximate bit error rate (BER) of MSK.
[A]:
A. See lecture note
B. The average probability of symbol error is
๐๐ = 1 โ ๐๐ = erfc (โ๐ธ
2๐0) โ
1
4erfc2 (โ
๐ธ
2๐0)
๐๐ โ erfc (โ๐ธ
2๐0) = erfc (โ
๐ธ๐๐0) , when
๐ธ
2๐0โซ 1
For Gray coded symbols, the most probable number of bit errors is one as a
symbol is most likely to be taken as adjacent symbols. As a result
โด BER =๐๐2=1
2erfc (โ
๐ธ๐๐0)
C. For the optimum detection of ๐(0), we project x(t) onto ๐1(๐ก)
๐ฅ1 = โซ ๐ฅ(๐ก)๐1(๐ก)๐๐
โ๐๐
๐๐ก = โ๐ธ๐cos ๐(0) + ๐ค1, โ ๐๐ โค ๐ก โค ๐๐
If ๐ฅ1 > 0, then ๐(0) = 0, otherwise ๐(0) = ๐.
For the optimum detection of ๐(๐๐), we project x(t) onto ๐2(๐ก)
๐ฅ2 = โซ ๐ฅ(๐ก)๐2(๐ก)2๐๐
0
๐๐ก = โโ๐ธ๐sin ๐(๐๐) + ๐ค2, 0 โค ๐ก โค 2๐๐
If ๐ฅ2 > 0, then ๐(๐๐) = โ๐
2, otherwise ๐(๐๐) =
๐
2.
D. The average correction I s
๐๐ = (1 โ ๐โฒ)2 = 1 โ erfc (โ๐ธ๐๐0) +
1
4erfc2 (โ
๐ธ๐๐0)
Thus the probability of wrong decision for one message points is โ
erfc (โ๐ธ๐
๐0) which yields the BER of ๐๐ =
1
2erfc (โ
๐ธ๐
๐0)
5. (10+5+5=20 points)
For M-ary QAM with square constellations, i.e. M=4, 16, 64,โฆ, suppose the
energy of the signal with the lowest amplitude is E0,
A. Derive the symbol error rate of M-ary QAM.
B. Define the bandwidth efficiency as Rb/B, where Rb = 1/Tb is the bit rate, and
B=2/T with T being the symbol duration. Show the bandwidth efficiency of
M-ary QAM.
C. Compare the bandwidth efficiency of M-ary QAM with that of M-ary FSK
supposed that the adjacent frequencies of M-ary FSK are separated from
each other by 1/2T
[A]:
A. See the lecture for the details. Given the average symbol energy
๐ธ๐๐ฃ =2(๐ฟ2 โ 1)๐ธ0
3=2(๐ โ 1)๐ธ0
3
The average symbol error rate is
๐๐ โ 2(1 โ1
โ๐) erfc (โ
๐ธ0๐0) = 2(1 โ
1
โ๐) erfc (โ
3๐ธ๐๐ฃ2(๐ โ 1)๐0
)
B. The symbol duration T of M-ary QAM is ๐๐ log2๐. Since ๐ ๐ =1
๐๐, and B =
2
T=
2๐ ๐
log2๐, then ๐ =
๐ ๐
๐ต=
log2๐
2
C. For M-ary FSK, the bandwidth is B = M / 2T , then ๐ =๐ ๐
๐ต=
2๐๐๐2๐
๐
6. (5+5+5=15 points)
Given that the bit energy is Eb and the noise density is N0,
A. Show the baseband power spectral density (PSD) of the BFSK signal
B. Show the baseband PSD of the MSK signal
C. Compare the baseband PSDs with that of the QPSK signal
Notice that sinc (๐ฅ) =sin (๐๐ฅ)
๐ฅ
[A]:
A. The in-phase component of BFSK is completely independent of the input
binary wave, and is equal to โ2๐ธ๐/๐๐ cos(๐๐ก/๐๐). The PSD of the in-phase
component is, thus, equal to
๐ธ๐2๐๐
[๐ฟ (๐ โ1
2๐๐) + ๐ฟ (๐ +
1
2๐๐)]
Depending on the input value, the quadrature component is equal to g(t) or
-g(t) where
g(๐ก) = {โ2๐ธ๐๐๐
sin (๐๐ก
๐๐) , 0 โค ๐ก โค ๐๐
0, otherwise
The ESD of g(t) is obtained by F(g(t)) F*(g(t)), yielding
๐นg(๐) =8๐ธ๐๐๐cos
2(๐๐๐๐)
๐2(4๐๐2๐2 โ 1)
2
The PSD thus equals to
๐๐ต(๐) =๐ธ๐2๐๐
[๐ฟ (๐ โ1
2๐๐) + ๐ฟ (๐ +
1
2๐๐)] +
8๐ธ๐cos2(๐๐๐๐)
๐2(4๐๐2๐2 โ 1)
2
B. Depending on the value of ๐(0), the in-phase component equals g(t) or -g(t)
where
g(๐ก) = {โ2๐ธ๐๐๐
cos (๐๐ก
2๐๐) , โ๐๐ โค ๐ก โค ๐๐
0, otherwise
The ESD of g(t) is obtained by F(g(t)) F*(g(t)), yielding
๐นg(๐) =32๐ธ๐๐๐cos
2(2๐๐๐๐)
๐2(16๐๐2๐2 โ 1)
2
The PSD equals ๐นg(๐)/2๐๐
Similarly, depending on the value of ๐(๐๐), the quadrature component is
equal to g(t) or โg(t) where
g(๐ก) = {โ2๐ธ๐๐๐
sin (๐๐ก
2๐๐) , 0 โค ๐ก โค 2๐๐
0, otherwise
The PSD is the same to that of the in-phase component
Since the in-phase and the quadrature components of MSK are statically
independent, the baseband PSD of the MSK equals
๐B(๐) = 2๐นg(๐)
2๐๐=
32๐ธ๐cos2(2๐๐๐๐)
๐2(16๐๐2๐2 โ 1)
2
C. For QPSK, the in-phase and quadrature components are statically
independent. The baseband PSD is thus equal to
๐๐ต(๐) = 2๐ธsinc2 (๐๐) = 2๐ธ๐ ๐๐2 (๐๐๐๐)
(๐๐๐๐)2= 2๐ธ๐ ๐๐๐2๐๐ ๐๐๐2 (๐๐๐ ๐๐๐2๐)
with M=4.
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